Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.1 Systems of Linear Equations: Substitution and Elimination - 10.1 Assess Your Understanding - Page 734: 34


$x=2, y = -3 \text{ or } (2, -3)$

Work Step by Step

The system of equations is given by the following two equations: $$\begin{cases} \dfrac{1}{2}x+y&=-2\hspace{20pt} &(1)\\[3mm] x-2y&=8 \hspace{20pt} &(2) \end{cases}$$ $\text{Multiplying equation (1) by 2 so that the coefficients of y in the two equations are additive inverses gives:}$ $x+2y=-4\hspace{20pt} (1)$ $\text{Add the two equations to eliminate } y:$ \[ \begin{aligned} &x+2y=-4 \\[2mm] &x-2y=8 \ \\[3mm] \hline & 2x=4\\[2mm] \end{aligned} \] Divide $2$ to both sides to obtain: $x=2$ $\text{Substituting in equation (1) for }x=2$ gives: $\dfrac{1}{2} (2)+y=-2$ $y=-2-1$ $y=-3$ Therefore, the solution is: $\boxed{x=2 \hspace{20pt} y = -3}$
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